Solve The Problem Graph Math
Given that the next problem is question 13 on page 411.
13. The figure above shows the graph of y=g(x). If the function h is defined by h(x)=g(2x)+2. What is the value of h(1)?
We looking for h(1)…
The first information is we can show the graph of y=g(x) in the co-ordinate system.
The next information is h(x)=g(2x)+2. We looking for h(1), so we can substitute x=1 on the equation, so h(1)=g(2)+2. Now we have g(2), we can look on the graph, if x=2 so y=1, so we can get g(2)=1.
The another function with no function problem, this is question 13, the another question 13 on page 534.
13. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
We looking for f(3p)… What is f when x = 3p?
Now first information is f(x)=x+1, and then the next piece information is 2f(p)=20, because the question is what is f when x = 3p, so we can start from this equation down here: 2f(p)=20, we can divide by 2, so f(p)=10, and then f(p) is just what is function f(x) of f(p),so we can speak f(p)=p+1=10, then p=9. But this is not the right answer, we looking for f when x = 3p, so if we take this and answer from x = 3p,with p=9, so x=27. Then we can substitute x in the equation f(x)=x+1, so f(27)=27+8=28. And the last answer of looking for is 28.
The next question is function with no function problem, we just take it step by step, this is question 17 on page 412.
17. In the xy–coordinate plane, the graph of x equals y square minus 4 intersect line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?
We will be looking for greatest m…
The graph x equals y square minus 4 intersect line l at (0,p) and (5,t), when x=0, y=p and when x = 5, y=t. And the question is what is the greatest slope. So what do we know about the slope? We know that m equals (y-y1)/(x2-x1). The slope is going to be m=(t-p)/5. Numerator is t-p. We can play again the value into this equation intersect at (0,p) and (5,t).
One way to find factors of polynomials is to formed the algebraic long division. For example lets try and see (x-3) is a factor of (x cube minus seven x minus six)? When dividing (x-3) into (x cube minus seven x minus six). First step of the problem make a long division problem for elementary school. There is you dividing (x – 3) into (x cube plus zero x square minus seven x minus six). Zero in there because is no second degree term. Now you must ask yourself what times x give you x cube? Of, course it is x square, and then you multiply (x-3) by x square, which give you (x cube minus 3 x square),you subtract (x cube plus zero x square) to get 3 x square. Bring it down to next term negative seven x, you have (3 x square minus 7 x). Now we begin again, dividing (x-3) into (3 x square minus 7 x). Just looking at the first term 3 x square dividing x is 3x. Multiply (x-3) by 3x. We can get (3 x square minus 9 x). Subtracting you have (2x minus 6). Dividing (x-3) into (2 x minus 6) which equals 2 and without a remainder. So the solution for a long division problem (x cube plus zero x square minus seven x minus six) dividing by (x-3) is (x square plus 3 x plus 2). Since (x-3) divide into (x cube minus 7 x minus 6) with no remainder. Then (x-3) is a factor of (x cube minus 7 x minus 6). The conclusion is (x square plus 3 x plus 2) is also a factor of (x cube minus 7 x minus 6). We know now that (x cube minus 7 x minus 6) equals (x-3) times (x square plus 3 x plus 2). The quadratic expression (x square plus 3 plus 2) can be factored into (x+1) times (x+2). So, (x cube minus 7 x minus 6) equals (x-3) times (x+1) times (x+2). Substitution (x cube minus 7x minus 6) to zero,so we get 0=(x-3)(x+1)(x+2) Thus either (x-3)=0 or (x+1)=0 or (x +2)=0. Solving all this equation of x we get x=3, x=-1, x=-2. The roots of (x cube minus 7 x minus 6) are (3, -1, -2).
¨ 3rd degree equation have 3 roots
¨ Quadratic (2nd degree) equations always have at most 2 roots
¨ A 4th degree equation would have 4 or fewer roots and so on.
¨ The degree of polynomials equation always limits the number of roots.
Long division process for 3rd order polynomial:
1. Find a partial quotient of x square, by dividing x into x cube to get x square.
2. Multiply x square by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “clear it out“ or reach a remainder.
Graph of A Rational Function
Graph of a rational function which can have discontinuities because has polynomial in the denominator.
Is possible value x divide by 0
f(x) equals (x+2) over (x-1)
when x=1,so f(1)= (1+2) over (1-1) equals 3 over zero. That is bad idea.
f(1)=(1+2) over zero is break in function graph.
f(x) = ( x+ 2) over (x-1)
for x=0, f(0)=(0+2) over (0-1) equals -2
insert x=1,so f(1)=1+2 over (1-1) equals 3 over 0, it is impossible.
Rational functions don’t always work in this way! Take graph f(x) = 1 over ( x square plus 1 ). Not all rational functions will give zero in denominator because of the (+1) is never zero.
Rational functions denominator can be zero!
Polynomial have smooth and unbroken curve and for rational function x : zero in the denominator that impossible situation. There is no value for the function.
A break can show up in 2 ways. A simply type break is missing point on the graph.
y= (x square minus x minus 6) over (x minus 3)
The graph loose like this if x = 3, so (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0. That is not possible, not feasible, and not allowed.
So that is no way if x = 3. This is a typical example to the missing point syndrome.
y = (3 square minus 3 minus 6) over (3 minus 3) equals 0 over 0
When you see result of 0 over 0 and also tell you direction be possible factor top and bottom of rational function and simplify.
y = (x square minus x minus 6) over (x minus 3) equals (x minus 3) times (x plus 2) all over (x minus 3) equals (x plus 2).
F(x,y) = 0
Function y = f(x) : VLT
1.1 function x=g(y) : HLT : invertible
Look at the graph of function y equals x square
Look at the graph function!
This is a equation, so:
We have function
What is the value of f(g(x))?
=x+1-1 = x
What is the value of g(f(x))?
=x-(1/2)+(1/2) = x
@ x=0, y=-1
V Asym x=1
H Asym @ y=-2
Look at the graph function!